Answer:
Yes . Frictions was not included for the calculation for acceleration .
Explanation:
This is because we're considering the weight of an object which isn't acceleration i:e frictionless.
If friction is included, the equation for acceleration becomes.
F= ma,(Newton second law of motion)
where F=frictional force , m =mass of the object and a=acceleration..
So therefore, a = F/m.
Answer:
W = 270.9 J
Explanation:
given,
F(x) = (12.9 N/m²) x²
work = Force x displacement
dW = F .dx
the push-rod moves from x₁= 1 m to x₂ = 4 m
integrating the above
![W = 12.9\times [\dfrac{x^3}{3}]_1^4 dx](https://tex.z-dn.net/?f=W%20%3D%2012.9%5Ctimes%20%5B%5Cdfrac%7Bx%5E3%7D%7B3%7D%5D_1%5E4%20dx)
`![W = 12.9\times [\dfrac{4^3}{3}-\dfrac{1^3}{3}] dx](https://tex.z-dn.net/?f=W%20%3D%2012.9%5Ctimes%20%5B%5Cdfrac%7B4%5E3%7D%7B3%7D-%5Cdfrac%7B1%5E3%7D%7B3%7D%5D%20dx)
W = 270.9 J
work done by the motor is W = 270.9 J
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